If the length of each of two equal sides of an isosceles triangle is 10 cm and the angle made by these side on the third side is 45°, then  find perimeter of the triangle.

10(2 + \(\sqrt {2}\)) cm

The triangle ABC is an isosceles triangle,

\(\angle\)BAC = \(\angle\)BCA = 45°

Therefore, \(\angle\)B = 90°

In ΔABC:

⇒ AC² = AB² + BC²

⇒ AC² = 100 + 100 = 200 cm

⇒ AC = 10 \(\sqrt {2}\) cm

Now,

Perimeter = 10 + 10 + 10\(\sqrt {2}\)

                = 10 (2 + \(\sqrt {2}\)cm)

                = 10(2 + \(\sqrt {2}\)) cm